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Rutherford's equation is called the Rutherford formula: N(θ) = N ^{2}k^{2}e^{4}/4r^{2}K^{2}sin^{4}(θ
/2) This equation is arrived at by treating the scattering as due to the Coulomb force, with the nucleus as a point-charge Z I have discovered that the math used to analyze scattering is incomplete. I say incomplete rather than false because it is correct as far as it goes. It simply fails to take into account the presence of gravity at the atomic level. This means that although its manipulations are done correctly, its assumptions are faulty. Rutherford assumes that the force can be expressed as the Coulomb force, and that therefore it is solely an electrostatic force. I will show that this is false. The intent of this paper is not to return to the Thomson model or to throw into question the usefulness of the Coulomb equation. As I admit above, the Coulomb equation allowed Rutherford to estimate the correct answer, so it must be correct as a heuristic device over some energies, at the very least. I am also not intending to question the experimental findings of the last century. I accept that the energies and forces have been measured correctly and that the data has been read properly. I only intend to show that the force field in this scattering problem is not the single field it has been assumed to be for a century. To do this I will apply my new Unified Field to the problem of scattering and show, using that field, that we must re-assess all our size estimates at the atomic and quantum level. In a series of papers I have shown that the charge field is present in Newton's gravitational equation F = GMm/r Unlike others who have tinkered with Newton's equation, I have not added the charge field as an extra term in the equation, or done anything that would change the current predictions of the equation. Rather, I have inserted the charge field into the existing terms, so that Newton's equation may stand as-is. I have only re-expanded the equation, so that we may see clearly the fields that lay under it and generate it. This makes Newton's equation a compound equation, with G as the transform between the two fields it contains. I have shown that the two fields are in vector opposition and that Newton's equation gives us only the field differential. That is, his equation tells us the total field but tells us nothing of the constituent fields. We can discover the relative size of these fields only by studying the interactions of real bodies. I have done this starting with the Earth and Moon and now have firm numbers not only for the relative field strengths of these two bodies, but for the relative field strengths in general. This discovery has also allowed me to explain the shape of the ellipse, the cause of orbital stability, the cause of planetary torques, and many other things. A major implication of this expansion of Newton's equation is that gravity is a function of radius and nothing else. If we write the masses in Newton's equation as density times volume, the density describes a quality of the charge field (or what I call the foundational E/M field) and has nothing to do with gravity at all. Gravity is shown to be an acceleration only, and whether that acceleration vector points in or out from the center it has nothing at all to do with density. This being so, the size of the field must change at a different rate, depending on the size of the gravitating object. In the end, this means that the size of the gravitational field at the quantum level must be some 10 This also gives us an easy way to calculate gravity in QED. If the charge field is present in Newton's equation, then the gravitational field must be present in Coulomb's equation. I have shown that the two equations are really the same equation, with one hiding the charge field and the other hiding the gravitational field. When we re-expand Newton's equation, we find the charge field: F = GMm/r When we re-expand Coulomb's equation, we find the gravitational field: F = kQq/r H is the gravitational field and it is found by the same equation at all levels of size. The variable "m" is the mass of the smaller of the two objects, being the gravitating object in Newton's equation and the scattered particle in Rutherford's experiment. "A" is the gravitational acceleration of the larger object and "a" of the smaller. As you see, this must impact the findings of Rutherford and all scattering experiments. It will not change the data, of course, but it must change the mechanical assumptions. If the force is not only electrostatic, then the mechanics cannot be what we have assumed it is. If we mis-assign forces, we end up with wrong numbers when we start calculating down from those forces. In these scattering equations, we are not calculating energies or forces from lengths or times, we are calculating lengths or times from energies or forces. Logically, that is upside down, and it is a dangerous mathematical manipulation. It requires an assumption of complete knowledge of the field mechanics, so that we can solve down in the correct way. Since I have shown that we do not have a complete knowledge of the field mechanics, we should not be able to solve down with such complete assurance. In fact, Rutherford and those who followed him have solved down using false assumptions, and have thereby gotten the wrong numbers for their lengths. To find the diameter of the nucleus, current math seeks what is called the "impact parameter", which is the perpendicular distance to closest approach during scattering. Once the scattering departs from the Rutherford formula, it is assumed that this impact parameter is equal to the nuclear radius, since the b = √
[1 + cosθ
/1 - cosθ
] kQq/mv
As you can see, the impact parameter is calculated from both the Coulomb force and the kinetic energy.
In the numerator we have the constant k transforming the charges into usable numbers. Let me tell you what I mean by that. In Newton's equation, G has the job of turning the masses into usable numbers: numbers that will give us the correct force. We need a constant or a transform like G because the mass variables are not really standing for a single measurable field parameter. As I have shown, the mass variables in Newton's equation are standing for two separate qualities in two separate fields. The mass variable should be written as DV (density times volume), and we should give the D variable to the E/M field and the V variable to the solo gravitational field. Since we don't do this, and since we have had no knowledge of the two fields involved, we have simply let G make all the corrections for us. G turns our misunderstood and mis-defined mass variables into numbers that work. The constant k does the very same thing in Coulomb's equation. Charge has never been defined mechanically, so the charge variables here are physically meaningless. They are little more than place fillers. We have never given "charge" a definition, not even to the extent we have given "mass" a definition. The definition of "mass" is pretty slippery, from a mechanical point of view, but the definition of "charge" is even more slippery. Charge is now defined as a trading of virtual particles, which is about as non-mechanical as you can get. No, all the mathematical content is in the constant k. It does all the numerical work in Coulomb's equation. It is quite easy to see this, since the value of charge is just 1 or -1. The number "1" does nothing in an equation, even when it is given an exponent. These charge variables are just ghosts. We could remove them from the equation altogether and it wouldn't matter. In Coulomb's equation, the constant k expresses the unified field directly, compressing both the gravitational field and the E/M field into a single number that gives us the right force at this level of size. But it does even more work than G does in Newton's equation, since it not only acts as a transform between the two fields, it also acts to transform those two 1's into real numbers. At least in Newton's equation the mass variables have given numbers. We don't give every body in the universe a mass of 1, do we? But in Coulomb's equation we give every important body a charge of 1. Pretty silly, if you think about it. Anyway, k is basically transforming radius and density once again, the radius giving us the gravitational field and the density giving us the density of the This being so, we have a problem in the impact parameter equation, since we have transformed the charges into usable numbers with the constant k, but we have not transformed the mass into a usable number. In this equation, we aren't just dealing with kinetic energies, we are dealing with compound force fields. Once you insert a mass or a kinetic energy into a field, you need to transform it so that it gives the right number. But here we have a mass being used in a field equation without the transform G, k, or any other transform. You will say, I see your point, but how on earth are you going to develop a transform like that? You can't use k as the transform, since k transforms charges, taking them from the number 1 to a usable number. You can't use G, since G transforms two intersecting fields of two bodies into one number, given two masses. Here you just have one mass, entering fields that have already been transformed in the numerator by k. It does look like a job, but let's tackle it. Let us start with an expression of k that isn't used very much: k = 10 That c E = mc Setting the two c E/m = k/10 m = 10 In the Rutherford experiment, departure from the Rutherford formula began at about 27.5MeV. 1eV = 1.6 x 10 m = 10 m = 4.4 x 10 According to that equation, the mass in our field equation should be m = 4.4 x 10 But we already know that the mass of our _{m} = m_{k}/m_{α}
= .00736
b = √
[1 + cosθ
/1 - cosθ
] kQq/η
We needed to find a transform for our mass, and we have done that. I think you may be surprised how little work it actually took. There are longer and more complex ways to derive this transform, but this is by far the most elegant. There are other ways we could write that last equation, too. We could just write it in terms of m b = √
[1 + cosθ
/1 - cosθ
] kQq/m Or we can take the mass out of the equation altogether b = √
[1 + cosθ
/1 - cosθ
] k Where ξ is a new constant ξ = 4.4 x 10^{-19 }J
Perhaps not surprisingly, that is the energy of a photon. We multiply our mass m 1/.00736 = 136, so our new value for b must be about 136 times the current value. Which means that the nucleus is 136 times larger than the current estimate. You may want to compare that number to the fine structure constant. I just showed that you have to multiply a mass by .00736 to turn it into charge, at the quantum level. The fine structure constant is about .0073. Therefore, I have just showed you where the fine structure constant comes from, answering one of Feynman's top questions, a question he put on his blackboard every morning for 20 years. Everything in the atomic and quantum world—including the atom, the nucleus, the proton, and the electron—is about 100 times larger than you were taught. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |